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63

A sonic boom is produced when a macroscopic object (say, roughly: larger than the average spacing between air molecules, $\approx 3\,\mathrm{nm}$) moves so fast that the air has no time to “get out of its way” in the usual way (linearly responding1 to a pressure buildup, which creates a normal sound wave that disperses rather quickly, more or ...


39

I know that when an object exceeds the speed of sound[340 m/s] a asonic boom is produced .Light which travels at 300000000m/s [much more than the speed of sound] doesn't produce a sonic boom right? Why? The answer is already in your own question: just because light is not an object. Sound "is a vibration that propagates as a typically audible ...


21

There are many differences between light and sound waves noted in other answers, such as the impossibility of any object with nonzero rest mass reaching lightspeed. However, there is one likeness that I don't think has been noticed yet and that is the following: a sound wave travelling at the speed of sound does not make a sonic boom! This is because the ...


6

First: what frequency should you hit? There are many, many different factors at play in determining the natural frequency of an object I know from experience. These are (not limited to): Thickness, density, elasticity modulus (you'll need two of those, e.g. Young's Modulus and Poisson Ratio), and of course shape. I'm not aware of any papers publishing a ...


5

There are two primary factors that allow the cochlea to isolate frequencies. These are generally referred to as passive and active properties: tl;dr version: The passive properties are due to the mechnical properties of one of the membranes in the cochlea, the basilar membrane, primarily the width and stiffness at a given point. The active properties are ...


5

Internal friction in the metal of the bell eventually will bring the ringing vibrations to an end. The bell vibrates when it rings, making its molecules more energetic and creating heat. Bonding between the molecules of the bell resist the vibrations, and eventually the strength of the molecular bonds will create enough friction to bring the vibrations ...


4

Anything that "suspends" the bell - whether it be a bolt, a piece of string, or a magnetic field - is applying a force. When the bell vibrates, this vibration will be transmitted. This is because the force of a magnet is a function of position - you can only get magnetic attraction because of a divergence of the field, so if you move, the force changes and ...


4

I find the explanation given in the first paragraph of Wikipedia article is pretty good. Let me just elaborate some aspects to make it more clear. Megaphone is simply an extension of your vocal tract. Therefore the acoustic impedance of the whole system rises so the pressure and volume flow variations at your vocal chords may grove. A trade-off is ...


3

When one shouts , the sound waves disperse in a semicircle , the power of the voice cords distributed to 180 degrees. A simple megaphone channels the sound in a small angle and thus is directional and stronger. Electric megaphones amplify the sound and still send it in a narrow cone. At the output of the cone the sound wave spreads, but it still is much ...


3

To the point of Is it hard to measure the resonant frequencies directly: it's tricky and careful discussion of the measuring procedures is needed. Some of the main problems: Destruction of the open-end behavior: If you place the speaker and microphone in front of the vocal tract to measure the response, you may have just switched open end behavior of your ...


3

There are just two requirements, 1) correct frequency, and 2) sufficient amplitude. The correct frequency is, the resonant frequency of the glass cup (pane, cube, etc.). You will know you have sufficient amplitude, when the glass breaks! Both requirements will vary, depending on the material, shape, dimensions of the object, and other variables. If you ...


3

The resonances are quite broad: each cavity will amplify a broad range of frequencies, spanning most of or more than an octave. Driving those resonances isn't as simple as choosing a pitch. You have to do some work to efficiently couple the different cavities to your vocal apparatus, and to maintain the resonance while you're singing. The people who are ...


2

I think The Physics of Musical Instruments (Springer Science & Business Media, 1998) by Fletcher and Rossing would be a good starting point for you. The general physical description of sound rests on the investigation of the impedance changes on the boundaries. For example: the reflection at the end of the string is caused by the discontinuity between ...


2

Yes sound is a goldstone mode. Consider, for example, an ideal gas with particles at positions $\mathbf{x}_i$. There is a symmetry where we can displace each particle by some displacement $\mathbf{u}$. Of course this symmetry breaks spontaneously. By definition, we only observe $\mathbf{u}=\mathbf{0}$. The goldstone modes corresponding to this symmetry are ...


2

The speed of sound is given by: $$v = \sqrt{\gamma\frac{P}{\rho}} \tag{1} $$ where $P$ is the pressure and $\rho$ is the density of the gas. $\gamma$ is a constant called the adiabatic index. The equation should make intuitive sense. The density is a measure of how heavy the gas is, and heavy things oscillate slower. The pressure is a measure of how stiff ...


2

Your error is that you assume it takes 2.00s seconds for the ping to reach the cliff and an additional 1.98s to return. Without knowing the distance to the cliff, we can't qualify that assumption. Besides that, the only thing we need for the calculation is the difference in period of the outgoing and incoming pings. With a single ping you can calculate the ...


2

The round trip time of the ping is unknown; but we do know that the difference in round trip time between sub stationary and sub moving is 0.02 seconds. Let us write $D$ for the distance to the cliff when you send the ping; if you are traveling at a speed $v$, and the speed of sound in water is $c$, then we can write down the round trip time as follows ...


1

I know this question is technically already answered, but there were several things missing from the answers that I thought should be mentioned (I am writing a review paper comparing different regions of space so I had these numbers at hand already as well). The speed of sound in space has multiple meanings because space is not a vacuum (though the number ...


1

Any structure that leads to a high Q system (the glass) will work and the trick is precisely matching the resonant (natural frequency ). By mounting the glass in a clamp that dissipates energy at a lesser rate than the sound energy that feeds it, the glass is doomed regardless of thickness or lack of imperfections. If the rate of energy input exceeds the ...


1

There is a model described in Main's Vibrations and Waves in Physics dealing with the speed of sound variations you might consider useful. Sorry, I would just comment that, but I don't have enough reputation. The other way might be to derive the speed of sound not from the ideal gas laws but from van der Waals equation, but to be honest, I've never tried ...


1

There are distance measurement devices commercially available, but if none will do, then I recommend making your own using a pulsed laser, a detector, and accompanying electronics. Another method would be a small buoy with a radio transmitter and a sound generator inside. The radio transmitter sends a signal once per minute and the sound generator emits a ...


1

Yes. a wave created with a certain frequency in the water remains the same even if the medium is changed. The only thing that is important in here is the amplitude of our wave. When the sound wave from inside of water hit the surface, some of the wave reflect back into the water and a lower amplitude wave is continued in the air. so the energy of the wave ...


1

I am reasoning my way to an answer here - I have never seen this problem before so I could be completely wrong. I think the issue is that you need to make sure the normal force is reduced as the tension on the "signal" side of the capstan is reduced. This is presumably why the slightly stiffer wire (as opposed to the rope) gave more predictable results at ...


1

well, that's good but not perfect. because in this model you can not have volume 0 unless you infinitely far from the source. the volume zero for us human should be when we can not hear any more sound. In physics it is called "threshold of hearing". so we want our function Volume to be some how that when we get pass the threshold of hearing, the volume ...


1

I know very little about sound and the ways to measure intensity and power, so please correct me if I confuse terms and concepts. You probably do not want to use intensity and power at all. Intensity is the product of particle velocity and sound pressure. To me it seems that pressure is the quantity you should use. Sound power is as power as any power ...


1

The musical property of a guitar, a violin, a cello, and indeed any string instrument, depend to a great extent on the shape of the empty space they contain. Such property also depends on the vibrational properties of the wood that encloses their empty space. Empty space properly enclosed within a thin skin is like an echo chamber that can magnify ...


1

The very famous Newton-Laplace equation is a relation between the speed of sound and the pressure of an ideal gas. It can be written as: $$ v = \sqrt{\gamma P / \rho} $$ where v is the velocity of sound in the given medium, P is the pressure, γ is the ratio of the heat capacities for the medium and ρ is the density of the medium. The Newton-Laplace was ...


1

Water forms close to perfect spheres in zero gravity due to it's surface tension. There's a variety of videos of water in the space station. Ice, assuming you start with one of those balls of water, you have to ask first, would it freeze outside in (say, the temperature of the station is dropped below 0 C), or would it freeze inside-out, say you stick a ...


1

Let's review the linearisation and go to the further details. Just the pressure might be not enough. Take the momentum equation: $$ -\frac{1}{\rho}\nabla p = \frac{\partial \vec{v}}{\partial t}+\vec{v}\cdot\nabla\vec{v} $$ Here we have to eliminate the convective part $\vec{v}\cdot\nabla\vec{v}$. Usually the argumentation is that changes of the velocity ...



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